Arjun Suresh (talk | contribs) |
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| − | Let G=(V, E) be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then | + | Let $G=(V, E)$ be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then |
| − | (A) | + | (A) $|S| = 2|T|$ |
| − | (B) |S| = |T| - 1 | + | (B) $|S| = |T| - 1$ |
| − | '''(C) |S| = |T| ''' | + | '''(C)$ |S| = |T| $''' |
| − | (D) |S| = |T| + 1 | + | (D) $|S| = |T| + 1$ |
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
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<br> | <br> | ||
<ul> | <ul> | ||
| − | <li>'''Base case''' : When $\xi(S) = \xi(T) = 1$, then we have only single node in both trees and hence |S| = |T|.</li> | + | <li>'''Base case''' : When $\xi(S) = \xi(T) = 1$, then we have only single node in both trees and hence $|S| = |T|$.</li> |
| − | <li>'''Induction hypothesis''' : Assume, for $\xi(S) = \xi(T) = k$, we have |S| = |T|.</li> | + | <li>'''Induction hypothesis''' : Assume, for $\xi(S) = \xi(T) = k$, we have $|S| = |T|$.</li> |
| − | <li>'''Induction step''' : For $\xi(S) = \xi(T) = k+1$, we must add a leaf vertex in both trees, and hence |S| = |T|.</li> | + | <li>'''Induction step''' : For $\xi(S) = \xi(T) = k+1$, we must add a leaf vertex in both trees, and hence $|S| = |T|$.</li> |
</ul> | </ul> | ||
| − | So |S| = |T|. So option (C) is correct. | + | So $|S| = |T|$. So option (C) is correct. |
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[[Category: GATE2010]] | [[Category: GATE2010]] | ||
| − | [[Category: Graph Theory questions | + | [[Category: Graph Theory questions from GATE]] |
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Let $G=(V, E)$ be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then
(A) $|S| = 2|T|$
(B) $|S| = |T| - 1$
(C)$ |S| = |T| $
(D) $|S| = |T| + 1$
$\xi(G)$ basically is the sum of degrees of all vertices in a graph. If sum of degrees of two different trees are same,
then number of nodes in the trees has to be same. We prove this by induction on sum of degree of all vertices in the two trees.
So $|S| = |T|$. So option (C) is correct.
Let G=(V, E) be a graph. Define $\xi(G) = \sum\limits_d i_d*d$, where $i_d$ is the number of vertices of degree d in G. If S and T are two different trees with $\xi(S) = \xi(T)$, then
(A) <|S| = 2|T|
(B) |S| = |T| - 1
(C) |S| = |T|
(D) |S| = |T| + 1
$\xi(G)$ basically is the sum of degrees of all vertices in a graph. If sum of degrees of two different trees are same,
then number of nodes in the trees has to be same. We prove this by induction on sum of degree of all vertices in the two trees.
So |S| = |T|. So option (C) is correct.
GATECSE